为什么' typeid(x)== typeid(y)'如果为' x'和' y'是类型T和T& amp; amp;的id表达式分别? [英] Why does 'typeid(x) == typeid(y)' evaluate to true, where 'x' and 'y' are id-expression of type T and T& respectively?
问题描述
我正在阅读C ++ 11草案标准，关于[expr.typeid]的部分提到了以下内容(强调我的意思):
I am reading the C++11 draft standard and the section on [expr.typeid] mentions the following (emphasis mine):
[...]
将typeid应用于多态类类型的glvalue以外的表达式时，结果将到表示表达式静态类型的std :: type_info对象.左值到右值(4.1)，数组指针(4.2)和函数到指针(4.3)的转换不适用于该表达式.如果类型表达式是一个类类型，该类必须是完全定义的.表达式是未计算的操作数(第5条).
When typeid is applied to an expression other than a glvalue of a polymorphic class type, the result refers to a std::type_info object representing the static type of the expression. Lvalue-to-rvalue (4.1), array-topointer (4.2), and function-to-pointer (4.3) conversions are not applied to the expression. If the type of the expression is a class type, the class shall be completely-defined. The expression is an unevaluated operand (Clause 5).
将typeid应用于 type-id 时，结果引用一个std :: type_info对象，该对象表示类型ID.如果type-id的类型是对可能由cv限定的类型的引用，则typeid的结果表达式引用了一个std :: type_info对象，该对象代表了不合格的引用类型.如果type-id是类类型或对类类型的引用，则该类应完全定义.
When typeid is applied to a type-id, the result refers to a std::type_info object representing the type of the type-id. If the type of the type-id is a reference to a possibly cv-qualified type, the result of the typeid expression refers to a std::type_info object representing the cv-unqualified referenced type. If the type of the type-id is a class type or a reference to a class type, the class shall be completely-defined.
在同一部分的p5中，进一步给出以下示例:
Further in p5 of the same section, it goes on to give the following example:
class D { /* ... */ };
D d1;
const D d2;
typeid(d1) == typeid(d2); // yields true
typeid(D) == typeid(const D); // yields true
typeid(D) == typeid(d2); // yields true
typeid(D) == typeid(const D&); // yields true -- (1)
给出以下代码示例:
Given the following code sample:
int main()
{
int foo = 42;
int &bar = foo;
bool comp1 = (typeid(int) == typeid(int&)); // Yields true, same as (1) -- (2)
bool comp2 = (typeid(foo) == typeid(bar)); // Yields true, Why? -- (3)
}
我的理解是 [expr.typeid] p4 仅讨论 typeid( type-id )和 bar的形式是 id-expression ，而不是 type-id .为什么上面的(3)评估为 typeid(bar)中的 true ?标准中的哪个文本涵盖了这一点?我错过了什么?
My understanding is that [expr.typeid]p4 talks only about the form typeid(type-id) and bar in typeid(bar) is an id-expression and not a type-id. Why does (3) above evaluate to true? Which text in the standard covers this? What have I missed?
推荐答案
答案在 [expr]
5 如果表达式最初的类型为对T"([dcl.ref]，[dcl.init.ref])，将类型先调整为T分析.该表达式指定由表示的对象或功能引用，并且表达式是左值还是右值，具体取决于在表达式上.
5 If an expression initially has the type "reference to T" ([dcl.ref], [dcl.init.ref]), the type is adjusted to T prior to any further analysis. The expression designates the object or function denoted by the reference, and the expression is an lvalue or an xvalue, depending on the expression.
因此，当我们进入 [expr.typeid]
3 当typeid应用于其他表达式时而不是一个多态类类型，结果引用一个std :: type_info对象表示表达式的静态类型.左值到右值([conv.lval])，数组到指针([conv.array])和函数到指针([conv.func])转换不适用于表达.如果表达式的类型是类类型，则该类应该是完全定义的.表达式是未计算的操作数(子句[expr]).
3 When typeid is applied to an expression other than a glvalue of a polymorphic class type, the result refers to a std::type_info object representing the static type of the expression. Lvalue-to-rvalue ([conv.lval]), array-to-pointer ([conv.array]), and function-to-pointer ([conv.func]) conversions are not applied to the expression. If the type of the expression is a class type, the class shall be completely-defined. The expression is an unevaluated operand (Clause [expr]).
当 typeid 检查它时，所讨论的id表达式已具有引用的类型.
The id-expression in question is already of the referenced type when typeid examines it.
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